Question

If
$(^{40}C_0) + (^{41}C_1) + (^{42}C_2) + ...... + (^{60}C_{20}) \frac{m}{n} ^{60}C_{20}$
m and n are coprime, then m + n is equal to _____.

Answer 1

The correct answer is 102
$^{40}C_0 \+ ^{41}C_1 \+ ^{42}C_2 \+ ...... + ^{60}C_{20}$
$= ^{40}C_{40} + ^{41}C_{40} + ^{42}C_{40} + ...... + ^{60}C_{40 }$
$= ^{61}C_{41}$
$= \frac{61}{41} . ^{60}C_{41}$
∴ m = 61 , _n _= 41
Therefore , m + n = 102